Problem: $f(x)= \begin{cases} \dfrac{ x + 1 }{ ( x + 1 )( x - 9 ) } & \text{if } x \neq 9 \\ 3 & \text{if } x = 9 \end{cases}$ What is the domain of the real-valued function $f(x)$ ?
Explanation: $f(x)$ is a piecewise function, so we need to examine where each piece is undefined. The first piecewise definition of $f(x)$ $\frac{ x + 1 }{ ( x + 1 )( x - 9 ) }$ , is undefined when its denominator is 0. The denominator is 0 when $x=-1$ or $x=9$ So, based on the first piecewise definition, we know that $x \neq -1$ and $x \neq 9$ However, the second piecewise definition applies when $x = 9$ , and the second piecewise definition, $3$ , has no weird gaps or holes, so $f(x)$ is defined at $x = 9$ So the only restriction on the domain is that $x \neq -1$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid x \neq-1\, \}$.